`arcsinx + arcsiny = pi/2 , (sqrt(2)/2, sqrt(2)/2)` Use implicit differentiation to find an equation of the tangent line at the given point

`arcsin(x) +arcsin(y) = pi/2`

First, take the derivative of both sides of the equation using implicit differentiation.

`d/dx[arcsin(x) + arcsin(y)] = d/dx (pi/2)`

`d/dx[arcsin(x)] + d/dx[arcsin(y)]=d/dx(pi/2)`

Take note that the derivative formula of arcsine is

• `d/dx[arcsin(u)] = 1/sqrt(1-u^2)*(du)/dx`

And the derivative of a constant is zero.

• `d/dx(c) = 0`

Applying these two formulas, the equation becomes

`1/sqrt(1-x^2) *d/dx(x) + 1/sqrt(1-y^2)*d/dx(y) = 0`

`1/sqrt(1-x^2) *1 + 1/sqrt(1-y^2)*(dy)/dx=0`

`1/sqrt(1-x^2) + 1/sqrt(1-y^2)*(dy)/dx = 0`

Then, isolate `(dy)/dx` .

`1/sqrt(1-y^2)*(dy)/dx = -1/sqrt(1-x^2)`

`(dy)/dx = -1/sqrt(1-x^2)*sqrt(1-y^2)/1`

`(dy)/dx = -sqrt(1-y^2)/sqrt(1-x^2)`

Then, plug-in the given point to get the slope of the curve on that point.  The given point is `(sqrt2/2,sqrt2/2)` .

`(dy)/dx = -sqrt(1- (sqrt2/2)^2)/sqrt(1-(sqrt2/2)^2)=-1`

Take note that the slope of a curve at point (x,y) is the slope of the line tangent to that point.  Hence, the slope of the tangent line is

`m = (dy)/(dx) = -1`

Now that the slope of line that is tangent to the graph of function at `(sqrt2/2,sqrt2/2)` is known, apply the point-slope form to get the equation of the line.

`y-y_1 = m(x- x_1)`

Plugging in the values, it becomes

`y - sqrt2/2=-1(x - sqrt2/2)`

`y-sqrt2/2=-x + sqrt2/2`

`y = -x+sqrt2/2+sqrt2/2`

`y = -x + (2sqrt2)/2`

`y = -x + sqrt2`

Therefore, the equation of the tangent line is `y = -x + sqrt2` .